/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 31 Suppose that \(0 \leq a_{n+1}0 ... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 31

Suppose that \(0 \leq a_{n+1}0 $$ where \(\left\\{w_{n}\right\\}\) is a sequence of positive numbers such that $$ \sum w_{n}\left(a_{n}-a_{n+1}\right)=\infty $$ Show that \(\sum a_{n} b_{n}=\infty .\) HINT: Use summation by parts.

Short Answer

Expert verified
Question: Prove that \(\sum a_n b_n = \infty\) under the given conditions. Answer: We applied Abel's sum transformation to rewrite the series and analyzed it using the given conditions. We found that the series indeed diverges to infinity.

Step by step solution

01

Rewrite the series as a summation by parts

We start by rewriting the series \(\sum a_nb_n\) as a summation by parts. To do this, we apply Abel's sum transformation as follows: $$ \sum a_nb_n = a_1b_1 + \sum_{n=1}^\infty (a_n - a_{n+1})(b_1 + b_2 + \cdots + b_n) $$ Now we need to analyze the series in this form and determine its limit.
02

Analyze the transformed series

Now we look at the second term in the transformed series, which is: $$ \sum_{n=1}^\infty (a_n - a_{n+1})(b_1 + b_2 + \cdots + b_n) $$ Recognizing that the given condition \(\sum w_n(a_n - a_{n+1}) = \infty\) is for the same type of series, we can rewrite the above series as: $$ \sum_{n=1}^\infty w_n(a_n - a_{n+1}) \frac{b_1 + b_2 + \cdots + b_n}{w_n} $$ We know that the series \(\sum w_n(a_n - a_{n+1})\) diverges to infinity, so we focus on the second part, \(\frac{b_1 + b_2 + \cdots + b_n}{w_n}\).
03

Determine the limit of the series

Let's recall the second condition, which is: $$ \varliminf_{n \rightarrow \infty} \frac{b_{1}+b_{2}+\cdots+b_{n}}{w_{n}} > 0 $$ This tells us that for sufficiently large \(n\), the ratio \(\frac{b_{1}+b_{2}+\cdots+b_{n}}{w_{n}}\) remains positive. Thus, when multiplying it with the diverging series \(\sum w_n(a_n - a_{n+1})\), the resulting series also diverges to infinity. Finally, we have: $$ \sum a_nb_n = a_1b_1 + \sum_{n=1}^\infty (a_n - a_{n+1})(b_1 + b_2 + \cdots + b_n) = \infty $$ We have shown that \(\sum a_n b_n = \infty\).

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Most popular questions from this chapter

Complete the proof of Theorem 4.5 .3 by showing that (a) \(R=0\) if \(\lim _{n \rightarrow \infty}\left|a_{n+1}\right| /\left|a_{n}\right|=\infty ;\) (b) \(R=\infty\) if \(\lim _{n \rightarrow \infty}\left|a_{n+1}\right| /\left|a_{n}\right|=0\).

Suppose that \(k\) is a positive integer and $$ f(x)=\sum_{n=0}^{\infty} a_{n} x^{n} $$ has radius of convergence \(R\). Show that the series $$ g(x)=f\left(x^{k}\right)=\sum_{n=0}^{\infty} a_{n} x^{k n} $$ has radius of convergence \(R^{1 / k}\).

Derive the identity $$ 2 \sin x \cos x=\sin 2 x $$ from the Maclaurin series for \(\sin x, \cos x,\) and \(\sin 2 x\).

Prove: If \(\left\\{F_{n}\right\\}\) converges to \(F\) on \([a, b]\) and \(F_{n}\) is nondecreasing for each \(n,\) then \(F\) is nondecreasing.

Use the integral test to find all values of \(p\) for which the series converges. (a) \(\sum \frac{n}{\left(n^{2}-1\right)^{p}}\) (b) \(\sum \frac{n^{2}}{\left(n^{3}+4\right)^{p}}\) (c) \(\sum \frac{\sinh n}{(\cosh n)^{p}}\)
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